MCQ
Function $f(x) = {{4{x^2} + 1} \over x}$ is decreasing for interval
- A$\left( {{{ - 1} \over 2},\,{1 \over 2}} \right)$
- ✓$\left[ {{1 \over 2},\, - {1 \over 2}} \right]$
- C$(-1, 1)$
- D$[1, -1]$
$\frac{d}{{dx}}f(x) = \frac{d}{{dx}}\left[ {4x + \frac{1}{x}} \right] $
$= 4 - \frac{1}{{{x^2}}}$
For extremum, $\frac{{dy}}{{dx}} = 0$
==> $4 - \frac{1}{{{x^2}}} = 0$
==> $x = \frac{1}{2},\, - \frac{1}{2}$
$f\;\left( {\frac{1}{2}} \right) = 4.\frac{1}{2} + \frac{1}{{1/2}}$ = $2 + 2 = 4$
$f\;\left( { - \frac{1}{2}} \right) = 4.\left( { - \frac{1}{2}} \right) + \frac{1}{{ - 1/2}} = - 2 - 2 = - 4$
Hence $f(x)$ is decreasing in interval $[1/2,\, - 1/2]$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| $I$ | $II$ | $III$ | $IV$ |
| $f'(x) = \frac{9}{{28}} x^{7/3} +9$ | $f (x) = \frac{9}{{28}} x^{7/3} -2$ | $f (x) = \frac{3}{{4}}\,x^{4/3} +6$ | $f'(x) =\frac{3}{{4}}\,x^{4/3} -4$ |
| Column $I$ | Column $II$ |
| $(A)$ The set $\left\{\operatorname{Re}\left(\frac{2 i z}{1-z^2}\right): z\right.$ is a complex number, $\left.|z|=1, z \neq \pm 1\right\}$ is | $(p)$ $(-\infty,-1) \cup(1, \infty)$ |
| $(B)$ The domain of the function $f(x)=\sin ^{-1}\left(\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\right)$ is is | $(q)$ $(-\infty, 0) \cup(0, \infty)$ |
| $(C)$ If $f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$, then the set $\left\{f(\theta): 0 \leq \theta<\frac{\pi}{2}\right\}$ is | $(r)$ $[2, \infty)$ |
| $(D)$ If $f(x)=x^{3 / 2}(3 x-10), x \geq 0$, then $f(x)$ is increasing in | $(s)$ $(-\infty,-1] \cup[1, \infty)$ |
| $(t)$ $(-\infty, 0] \cup[2, \infty)$ |